In regular polygon ABCDE..., we have ∠ACD = 120°. How many sides does the polygon have?

Question

In regular polygon ABCDE..., we have ∠ACD = 120°. How many sides does the polygon have?

How would I start with solving this? AC would have different lengths depending on the polygon.

Answer 1

Assume that there exist totally $n$ sides along the vertexes $D,E,\cdots,A$. Notice that $$120^o=\angle ACD \overset{m}=\frac{1}{2}\widehat{DE\cdots A}=\frac{1}{2}\cdot 360^o\cdot \frac{n}{n+3}.$$

Thus, $n=6.$ As a result, the number of the sides of the polygon is $n+3=9.$

Answer 2

The interior angle of a regular polygon is $$\angle ABC=\angle BCD=\frac{180(n-2)}{n}$$ Now the triangle ABC is isosceles, so $$\angle BCA = \frac12(180-\frac{180(n-2)}{n}) \\ \angle BCA+\angle ACD = \angle BCD \\ \frac12(180-\frac{180(n-2)}{n})+120=\frac{180(n-2)}{n}$$ This gives $n=9$.